Simplify; express your answer in exponential form. Assume $t\neq 0, q\neq 0$. $\dfrac{{(t^{2}q^{-2})^{-3}}}{{(t^{-5}q^{3})^{4}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(t^{2}q^{-2})^{-3} = (t^{2})^{-3}(q^{-2})^{-3}}$ On the left, we have ${t^{2}}$ to the exponent ${-3}$ . Now ${2 \times -3 = -6}$ , so ${(t^{2})^{-3} = t^{-6}}$ Apply the ideas above to simplify the equation. $\dfrac{{(t^{2}q^{-2})^{-3}}}{{(t^{-5}q^{3})^{4}}} = \dfrac{{t^{-6}q^{6}}}{{t^{-20}q^{12}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{-6}q^{6}}}{{t^{-20}q^{12}}} = \dfrac{{t^{-6}}}{{t^{-20}}} \cdot \dfrac{{q^{6}}}{{q^{12}}} = t^{{-6} - {(-20)}} \cdot q^{{6} - {12}} = t^{14}q^{-6}$